[proofplan] We first use the continuity of multiplication and inversion to show that the identity component is closed under products and inverses, hence is a subgroup. Then we use conjugation by an arbitrary element of $G$ to show that this subgroup is invariant under conjugation, so it is normal. Finally, we use elementary topology of manifolds: connected components are closed in every [topological space](/page/Topological%20Space), and are open in locally connected spaces. Left translation carries the identity component onto the component containing the translating element, which identifies all connected components with cosets of $G_0$. [/proofplan] [step:Use multiplication and inversion to prove that $G_0$ is a subgroup] Let \begin{align*} m:G\times G\to G \end{align*} be the multiplication map, defined by $m(x,y)=xy$, and let \begin{align*} \iota:G\to G \end{align*} be the inversion map, defined by $\iota(x)=x^{-1}$. Since $G$ is a Lie group, both maps are continuous. Because $G_0$ is connected and the [product of connected spaces is connected](/theorems/299), $G_0\times G_0$ is connected. Therefore $m(G_0\times G_0)$ is connected as the continuous image of a [connected space](/page/Connected%20Space). Since $e\in G_0$ and $m(e,e)=e$, the set $m(G_0\times G_0)$ is a connected subset of $G$ containing $e$. By maximality of the connected component $G_0$, we obtain \begin{align*} m(G_0\times G_0)\subset G_0. \end{align*} Thus $xy\in G_0$ for all $x,y\in G_0$. Similarly, $\iota(G_0)$ is connected as the continuous image of the connected space $G_0$. Since $\iota(e)=e$, the set $\iota(G_0)$ is a connected subset of $G$ containing $e$. Again by maximality of $G_0$, \begin{align*} \iota(G_0)\subset G_0. \end{align*} Thus $x^{-1}\in G_0$ for all $x\in G_0$. Since $e\in G_0$, closure under products and inverses proves that $G_0\le G$. [guided] The goal is to prove the subgroup axioms for $G_0$ using only connectedness and the continuity of the group operations. The identity element $e$ already belongs to $G_0$ because $G_0$ is, by definition, the connected component containing $e$. It remains to prove closure under multiplication and inversion. Define the multiplication map \begin{align*} m:G\times G\to G \end{align*} by $m(x,y)=xy$. Since $G$ is a Lie group, multiplication is continuous. The space $G_0$ is connected by definition of connected component, so the product space $G_0\times G_0$ is connected by the standard topological fact that a product of connected spaces is connected. Applying the standard fact that a [continuous image of a connected space is connected](/theorems/296), the subset $m(G_0\times G_0)\subset G$ is connected. It contains $e$, because $(e,e)\in G_0\times G_0$ and $m(e,e)=e$. Since $G_0$ is the maximal connected subset of $G$ containing $e$, every connected subset of $G$ containing $e$ must lie inside $G_0$. Hence \begin{align*} m(G_0\times G_0)\subset G_0. \end{align*} Equivalently, if $x,y\in G_0$, then $xy\in G_0$. Next define the inversion map \begin{align*} \iota:G\to G \end{align*} by $\iota(x)=x^{-1}$. This map is continuous because $G$ is a Lie group. Therefore $\iota(G_0)$ is connected. It contains $e$, since $\iota(e)=e$. By the same maximality argument for the connected component containing $e$, \begin{align*} \iota(G_0)\subset G_0. \end{align*} Thus $x^{-1}\in G_0$ whenever $x\in G_0$. We have shown that $e\in G_0$, that $G_0$ is closed under multiplication, and that $G_0$ is closed under inverses. Therefore $G_0$ is a subgroup of $G$. [/guided] [/step] [step:Use conjugation to prove normality] Fix $g\in G$. Define the conjugation map \begin{align*} C_g:G\to G \end{align*} by $C_g(x)=gxg^{-1}$. The map $C_g$ is a homeomorphism, with inverse $C_{g^{-1}}$, because it is a composition of left translation by $g$ and right translation by $g^{-1}$. Also $C_g(e)=e$. Since $C_g(G_0)$ is the continuous image of a connected space, it is connected. It contains $e$, so maximality of $G_0$ gives \begin{align*} C_g(G_0)\subset G_0. \end{align*} Applying the preceding connected-image argument to $C_{g^{-1}}$ gives \begin{align*} C_{g^{-1}}(G_0)\subset G_0. \end{align*} Since $C_g$ is the inverse of $C_{g^{-1}}$, applying $C_g$ to this inclusion gives \begin{align*} G_0\subset C_g(G_0). \end{align*} Therefore $C_g(G_0)=G_0$. Since $g\in G$ was arbitrary, $gG_0g^{-1}=G_0$ for every $g\in G$, so $G_0\trianglelefteq G$. [/step] [step:Prove that $G_0$ is closed and open] First, connected components are closed in any topological space. Indeed, if $C$ is a connected component of a topological space, then its closure $\overline{C}$ is connected: if $\overline{C}$ were separated by two disjoint nonempty relatively open sets, their intersections with the [dense subset](/page/Dense%20Subset) $C$ would separate $C$. Since $C$ is maximal connected, $\overline{C}\subset C$, and hence $\overline{C}=C$. Applying this to $C=G_0$ shows that $G_0$ is closed in $G$. Next, $G$ is a smooth manifold, hence is locally connected. In a locally connected space, every connected component is open: if $C$ is a connected component and $x\in C$, then there exists an open connected neighbourhood $U_x\subset G$ of $x$; the union $C\cup U_x$ is connected because $C\cap U_x$ contains $x$, so maximality of $C$ gives $U_x\subset C$. Therefore every $x\in C$ has an open neighbourhood contained in $C$, and $C$ is open. Applying this to $C=G_0$ shows that $G_0$ is open in $G$. [/step] [step:Identify every connected component as a coset of $G_0$] Fix $g\in G$, and let $K_g$ denote the connected component of $G$ containing $g$. Define the left translation map \begin{align*} L_g:G\to G \end{align*} by $L_g(x)=gx$. This map is a homeomorphism, with inverse $L_{g^{-1}}$. The image $L_g(G_0)=gG_0$ is connected because $G_0$ is connected and $L_g$ is continuous. It contains $g$, because $L_g(e)=g$. Hence \begin{align*} gG_0\subset K_g. \end{align*} Conversely, $L_{g^{-1}}(K_g)$ is connected and contains $e$. Therefore \begin{align*} L_{g^{-1}}(K_g)\subset G_0. \end{align*} Applying $L_g$ gives \begin{align*} K_g\subset gG_0. \end{align*} Thus \begin{align*} K_g=gG_0. \end{align*} Since $g\in G$ was arbitrary, every connected component of $G$ is a left coset of $G_0$. Because $G_0\trianglelefteq G$, each left coset $gG_0$ is also the right coset $G_0g$. This completes the proof. [/step]